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Chemistry

Chemistry. Class Exercise. Class Exercise - 1. Express the following numbers to three significant figures. (i) 6.022 × 10 23 (ii) 5.356 g (iii) 0.0652 g (iv) 13.230. Solution. 6.02 × 10 23 5.36 g 0.0652 g 13.2. Class Exercise - 2. What is the sum of 2.368 g and 1.02 g?. Solution.

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Chemistry

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  1. Chemistry

  2. Class Exercise

  3. Class Exercise - 1 Express the following numbers tothree significant figures.(i) 6.022 × 1023 (ii) 5.356 g(iii) 0.0652 g (iv) 13.230 Solution • 6.02 × 1023 • 5.36 g • 0.0652 g • 13.2

  4. Class Exercise - 2 What is the sum of 2.368 g and1.02 g? Solution = 3.39 g

  5. Class Exercise - 3 Express the result of the followingcalculation to the appropriate numberof significant figures816 × 0.02456 + 215.67 Solution 816 × 0.02456 = 20.0 Product rounded off to 3 significant figures becausethe least number of significant figure in thismultiplication is three. Rounded off to 235.7

  6. Class Exercise - 4 Solve the following calculations andexpress the results to appropriatenumber of significant figures.(i) 1.6 × 103 + 2.4 × 102 – 2.16 × 102(ii) Solution (i) 1.6 × 103 + .24 × 103 Rounded off to 1.8 × 103

  7. Class Exercise - 4 (ii) Rounded off to 1.6 × 103 or 16 × 102 = 7.525 × 103 (rounded off to 7.5 × 103)

  8. Class Exercise - 5 125 inches = 2.54 × 10-2 × 125 m Convert 10 feet 5 inches into SI unit. Solution 10 feet 5 inches = 125 inches 1 inch = 2.54 × 10-2 m = 317.5 × 10-2 m Rounded off to 317 × 10–2 m

  9. Class Exercise - 6 A football was observed to travel at a speedof 100 miles per hour. Express the speedin SI units. Solution 1 mile = 1.60 × 103 m 100 miles per hour = 4.4 × 10-4 × 105 m/s = 4.4 × 10 m/s = 44 m/s

  10. Class Exercise - 7 What do the following abbreviationsstand for? (i) O (ii) 2O (iii) O2 (iv) 3O2 Solution • Oxygen atom • 2 moles of oxygen atom • Oxygen molecule • 3 moles of oxygen molecule

  11. Class Exercise - 8 Among the substances given belowchoose the elements, mixtures andcompounds (i) Air (ii) Sand(iii) Diamond (iv) Brass Solution • Air - Mixture • Sand (SiO2) - Compound • Diamond (Carbon) - Element • Brass (Alloy of metal) - Mixture

  12. Class Exercise - 9 • Classify the following into elements • and compounds. • H2O • He • Cl2 • CO • Co Solution Element: He, Cl2, Co Compound: H2O and CO

  13. Class Exercise – 10 Explain the significance of the symbol H. Solution • Symbol H represents hydrogen element • Symbol H represents one atom of hydrogen atom • Symbol H also represents one mole of atoms, that is,6.023 × 1023 atoms of hydrogen. • Symbol H represents one gm of hydrogen.

  14. BASIC CONCEPTS OF CHEMISTRY Session - 2

  15. Session Opener

  16. Session Objectives

  17. Session Objectives • The law of conservation of mass • The law of definite proportions • The law of multiple proportions • The law of reciprocal proportions • Gay Lussac’s law of gaseous volumes • Dalton’s atomic theory • Modern atomic theory

  18. H2 + Cl2 2 HCl 2g 71g 73g Law of conservation of mass Total mass of the products remains equal to the total mass of the reactants.

  19. Question

  20. Illustrative Problem 8.4 g of sodium bicarbonate on reaction with 20.0 g of acetic acid (CH3COOH) liberated 4.4 g of carbon dioxide gas into atmosphere. What is the mass of residue left? Solution: 8.4 + 20.0 = m + 4.4 m = 24 g It proves the the law of conservation of mass.

  21. Law of definite proportions Ice water H2O 1 : 8 River water H2O 1 : 8 Sea water H2O 1 : 8 A chemical compound always contains same elements combined together in same proportion of mass.

  22. Question

  23. Illustrative Problem • Two gaseous samples were analyzed. • One contained 1.2g of carbon and • 3.2 g of oxygen. The other contained • 27.3 % carbon and 72.7% oxygen. The • experimental data are in accordance • with • Law of conservation of mass • Law of definite proportions • Law of multiple proportions • Law of reciprocal proportions

  24. Solution % of C in the 1st sample Which is same as in the second sample.Hence law of definite proportion is obeyed.

  25. Law of multiple proportions 14:8 14:16 14:24 14:32 14:40 Two elements combine two or more compounds Ratio of oxygen combining with 14 parts of nitrogen = 8:16:24:32:40 1:2:3:4:5 the mass of one of the elements which combines with fixed mass of the others, bear a simple whole number ratio to one another.

  26. Statement of law of reciprocalProportions The ratio of the weights of two elementsA and B which combine separately witha fixed weight of the third element ‘C’ iseither the same or some simple multipleof the ratio of the weights in which A andB combine directly with each other. k may be 1

  27. Law of reciprocal proportions H2S 2 : 32, 1 : 16 SO2 32 : 32, 1 : 1 S H S S O 2 2 H H O 2 O

  28. Gay Lussac’s law of gaseous volumes Gases reacts in volume which bear a simple ratio to one another and to the volume of the products under similar conditions of temperature and pressure.

  29. Gay Lussac’s Law of gaseous volumes 2 volumenitrogen oxide 1 volumenitogen gas 1 volumeoxygen gas + N2 O2 2NO 2 volumehydrogen gas 1 volumeoxygen gas 2 volumesteam + O2 2H2O 2H2 1 : 1 : 2 2 : 1 : 1

  30. Ask yourself Why Gay–Lussac’s law is not applicable to solids and liquids ? because they have negligible volumes as compared to gases.

  31. Do you know The laws of chemical combinations are basedon quantitative results of chemical reactions.

  32. Questions

  33. Illustrative Problem Carbon is found to form two oxides, which contains 42.8% and 27.27% of carbon respectively. Find out which of the laws of chemical combination is proved correct by this data? Solution: In the first oxide, Carbon :Oxygen = 0.428 : 0.572 = 0.748 : 1 In the second oxide, Carbon :Oxygen = 0.2727 : 0.7273 = 0.748 : 0.374 = 2:1 = 0.374 : 1

  34. Illustrative Problem Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Find out which of the laws of chemical combination is proved correct by this data? Solution: Hence k=1 which proves law of reciprocal proportion.

  35. Ask your self The balanced chemical reaction is an expression of Law of multiple proportion Law of conservation of mass Law of constant proportion None of the above

  36. Illustrative example Zinc sulphate crystals contain 22.6% of zinc and 43.6% of water. How muchZinc should be used to produce 13.7 gm of zinc sulphate crystals and how muchwater they will contain? (Hint: Law of constant composition) Solution: 100 gm of ZnSO4 will have 43.9 gm water and 22.6 gm zinc

  37. Solution

  38. Dalton’s Atomic Theory • Matter is made up of indivisible particles called atoms. • Atoms of the same element are similar with respect to shape, size and mass. • Atoms combine in simple whole number ratio to form molecule. • An atom can neither be created nor destroyed.

  39. Atoms of two elements may combine in different ratios to form more than one compound. SO2 1:2 S+O2 SO3 1:3

  40. Limitations of Dalton’s Atomic Theory It could not explain: Why atoms of different elementshave different masses, sizes, etc. Nature of binding force between atoms and molecules.

  41. Limitations of Dalton’s Atomic Theory • Can not explain causes of chemical combination • Can not explain law of combining volume • It does not give idea about structure of atom • It does not distinguish between the ultimate particleof an element and a compound • It does not give idea about isotopes and isobars.

  42. Modern Atomic Theory • Atom is divisible • Same atom may have differentatomic masses like 1H, 2H and 3H. • Different atoms may have same atomic mass like 40Ca and 40Ar. • Atom is the smallest particle that takes part in a chemical reaction. 5. The mass of an atom can be changed into energy.

  43. Class Test

  44. Class Exercise - 1 Percentage of copper and oxygen insamples of CuO obtained by differentmethods were found to be the same.This proves the law of (a) constant proportions (b) reciprocal proportions (c) multiple proportions (d) none of these Solution Definition

  45. Class Exercise - 2 A balanced chemical equation is inaccordance with (a) Boyle’s law(b) Avogadro’s law (c) Gay Lussac’s law(d) law of conservation of mass Solution Definition

  46. Class Exercise - 3 Different samples of water were foundto contain hydrogen and oxygen in theapproximate ratio of 1 : 8. This shows the law of (a) multiple proportion(b) constant proportion(c) reciprocal proportion(d) none of these Solution Definition

  47. Class Exercise - 4 Law of multiple proportion is illustratedby the following pair of compounds. (a) HCl and HNO3(b) KOH and KCl(c) N2O and NO(d) H2S and SO2 Solution Definition

  48. Class Exercise - 5 The oxides of nitrogen contain63.65%, 46.69% and 30.46%nitrogen respectively. These dataproves the law of (a) definite proportions (b) multiple proportions (c) reciprocal proportions (d) conservation of mass Solution Definition

  49. Class Exercise - 6 12 g carbon combine with 64 g sulphurto form CS2. 12 g of carbon alsocombine with 32 g oxygen to form CO2.10 g sulphur combine with 10 g oxygento form SO2. These data proves the (a) law of multiple proportions(b) law of definite proportions (c) law of reciprocal proportions(d) Cray Lussac’s Law of gaseous volume

  50. Solution Ratio of the weights of S and O combining with fixed weight of C is 64 : 32 = 2 : 1. Ratio of weights of S and O combining directly = 10 : 10 = 1 : 1. The two ratios are simple multiple of each other. This proves the law of reciprocal proportions.

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